# Reflection reverses a root string

I am trying to understand the part of the proof In Fulton's and Harris's Represantation Theory book where he shows that the length of the root string is at most 4:

Theorem If $$\alpha,\beta$$ are roots with $$\beta \neq \pm \alpha$$, then the $$\alpha$$ string through $$\beta$$, i.e. the roots of the form $$\begin{equation*} \beta - p\alpha, \beta -(p-1)\alpha, \dots, \beta - \alpha, \beta, \beta + \alpha, \beta + 2\alpha, \dots, \beta + q\alpha \end{equation*}$$ has at most four in a string, i.e. $$p+q \leq 3$$; in addition, $$p-q=n_{\beta\alpha}$$

where $$n_{\beta \alpha}= 2\frac{(\beta,\alpha)}{(\alpha,\alpha)}$$ (positive definite form on a vector space $$V$$) and then the reflection of a root $$\beta$$ in the hyperplane orthogonal to a root $$\alpha$$ is given by $$\begin{equation*} W_{\alpha}(\beta) = \beta - n_{\beta \alpha}\alpha \end{equation*}$$

The part of the proof I don't understand is when they write that $$$$W_{\alpha}(\beta + q\alpha) = \beta - p\alpha \quad (1)$$$$ meaning that the reflection (in the hyperplane orthogonal to $$\alpha$$) reverses a root string. I read somewhere else that this is geometrically an obvious fact but I can't see how. I get that by reflecting through $$\alpha$$ we just add a positive or negative multiple of $$\alpha$$ to any root but I really cannot see why $$(1)$$ is true.

• The elements of a root string all lie on an affine line with direction alpha (and they are the only points of the line in the root system). The points on this line furthest from the hyperplane orthogonal to alpha have to be mapped to one another via the reflection. Those furthest points are clearly the two ends of the root string. Is that geometrically obvious enough? Aug 20, 2019 at 18:13
• @SamHopkins Thank you! And yeah, it makes more sense now. Do you have any hints on how to show that algebraically? Aug 20, 2019 at 19:57
• I’m not sure of an essentially algebraic way to think about this: you’re going to have to use the fact that you have an inner product. Aug 20, 2019 at 20:32
• Yeah, I've been playing around with it (an inner product) but I haven't gotten anywhere yet. Anyway, thanks for your help! Aug 20, 2019 at 20:38
• Crossposted and answered on math.stackexchange: math.stackexchange.com/q/3328664/96384. Please note that crosspostings are not liked too much (and this question quite obviously belongs to stackexchange). Aug 23, 2019 at 3:39

EDIT (more details): Algebraically, the axioms show that the reflection $$s_\alpha$$ (written in GTM 9 as $$\sigma_\alpha$$) takes a non-proportional root $$\beta$$ to $$\beta + c \alpha$$ for some $$c \in \mathbb{Z}$$. Here the $$\alpha$$-string through $$\beta$$ is unbroken $$(*)$$. Thus the reflection $$s_\alpha$$ leaves the $$\alpha$$-string through $$\beta$$ invariant. To argue that $$s_\alpha$$ interchanges endpoints, we may assume that $$p+q \neq 0$$ (then $$s_\alpha$$ has order 2 on the string).
Now it is easiest to argue by cases. For example, if $$s_\alpha$$ takes $$\beta + q \alpha$$ to $$\beta - d \alpha$$ for some $$0 \leq d < p$$, then it must map this root back to the highest root in the string. In turn, this reflection must take the lowest element of the root string to a higher root than $$\beta + q \alpha$$, which is absurd. Etc.
[$$(*)$$ Note that the string is unbroken, which in Section 8 may require some representation theory of the rank 1 semisimple Lie algebra but in Section 9 (an axiomatic treatment of root systems) follows from the easy first step in the classification of root systems embodied in Table 1 amd used in the proof of Lemma 9.4. The upshot is that root strings (for the adjoint representation) have length at most 4, which is probably impossible to prove without taking a step into the classification. (By the way, the bound of 4 becomes unlimited when all irreducible finite dimensional representations are considered: this is the second item on my list Index of Terminology.)]